Question

47. The potential $V$ due to two infinite parallel filaments of charge of linear densities $\lambda$ and $-\lambda$ is $V = (\lambda/2\pi\epsilon_0)ln(r_2/r_1)$, where $r_1^2 = (x - x_0)^2 + y^2$ and $r_2^2 = (x + x_0)^2 + y^2$. We think of the filaments as being in the $z$ direction, passing through the $xy$ plane at $(-x_0, 0)$ and $(x_0, 0)$. (a) Find $\nabla V(x, y)$, using the chain rule. (b) Verify the flux law $\partial^2V/\partial x^2 + \partial^2V/\partial y^2 = 0$.

          47. The potential $V$ due to two infinite parallel filaments of charge of linear densities $\lambda$ and $-\lambda$ is
$V = (\lambda/2\pi\epsilon_0)ln(r_2/r_1)$, where $r_1^2 = (x - x_0)^2 + y^2$
and $r_2^2 = (x + x_0)^2 + y^2$. We think of the filaments as being in the $z$ direction, passing
through the $xy$ plane at $(-x_0, 0)$ and $(x_0, 0)$.
(a) Find $\nabla V(x, y)$, using the chain rule.
(b) Verify the flux law $\partial^2V/\partial x^2 + \partial^2V/\partial y^2 = 0$.

47. The potential V due to two infinite parallel filaments of charge of linear densities λ and -λ is
V = (λ/2πϵ0)ln(r2/r1), where r1^2 = (x - x0)^2 + y^2
and r2^2 = (x + x0)^2 + y^2. We think of the filaments as being in the z direction, passing
through the xy plane at (-x0, 0) and (x0, 0).
(a) Find ∇ V(x, y), using the chain rule.
(b) Verify the flux law ∂^2V/∂ x^2 + ∂^2V/∂ y^2 = 0.

Added by Megan G.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Supratim Pal

07/16/2023

Step 1

Using the chain rule, we have: Vx = λ * (1/(r/x0)) * (dr/dx) + λ * (1/(r/x0 + y)) * (dr/dx) Show more…

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47. The potential V due to two infinite parallel filaments of charge of linear densities λ and -λ is V = λ ln(r/x0) + λ ln(r/x0 + y). We think of the filaments as being in the z direction, passing through the xy plane at (-x0, 0) and (x0, 0). a) Find V (Vx, y), using the chain rule. b) Verify the flux law V/x + V/y = 0.